We consider termination of the LCSTRS with only rule scheme Calc:

  Signature: cons     :: A -> list -> list
             nil      :: list
             skip     :: Int -> list -> list
             skipfold :: (A -> B -> B) -> B -> Int -> list -> B
             zip      :: list -> list -> list

  Rules: skip(k, nil) -> nil | k = k
         skip(k, ys) -> ys | k <= 0
         skip(k, cons(x, ys)) -> skip(k - 1, ys) | k > 0
         skipfold(F, n, k, nil) -> n | k = k
         skipfold(F, n, k, cons(x, ys)) -> F(x, skipfold(F, n, k, skip(k - 1, ys))) | k = k
         zip(nil, ys) -> ys
         zip(cons(x, xs), ys) -> cons(x, zip(ys, xs))

The system is accessible function passing by a sort ordering that equates all sorts.

We start by computing the initial DP problem D1 = (P1, R, i, c), where:

  P1. (1) skip#(k, cons(x, ys)) => skip#(k - 1, ys) | k > 0
      (2) skipfold#(F, n, k, cons(x, ys)) => skip#(k - 1, ys) | k = k
      (3) skipfold#(F, n, k, cons(x, ys)) => skipfold#(F, n, k, skip(k - 1, ys)) | k = k
      (4) zip#(cons(x, xs), ys) => zip#(ys, xs)

***** We apply the Graph Processor on D1 = (P1, R, i, c).

We compute a graph approximation with the following edges:

  1: 1 
  2: 1 
  3: 2 3 
  4: 4 

There are 3 SCCs.

Processor output: { D2 = (P2, R, i, c) ; D3 = (P3, R, i, c) ; D4 = (P4, R, i, c) }, where:

  P2. (1) skip#(k, cons(x, ys)) => skip#(k - 1, ys) | k > 0

  P3. (1) skipfold#(F, n, k, cons(x, ys)) => skipfold#(F, n, k, skip(k - 1, ys)) | k = k

  P4. (1) zip#(cons(x, xs), ys) => zip#(ys, xs)

***** We apply the Subterm Criterion Processor on D2 = (P2, R, i, c).

We use the following projection function:

  nu(skip#) = 2

We thus have:

  (1) cons(x, ys) |>| ys

All DPs are strictly oriented, and may be removed.  Hence, this DP problem is finite.

Processor output: { }.

***** We apply the Usable Rules Processor on D3 = (P3, R, i, c).

We obtain 3 usable rules (out of 7 rules in the input problem).

Processor output: { D5 = (P3, R2, i, c) }, where:

  R2. (1) skip(k, nil) -> nil | k = k
      (2) skip(k, ys) -> ys | k <= 0
      (3) skip(k, cons(x, ys)) -> skip(k - 1, ys) | k > 0

***** We apply the Usable Rules Processor on D4 = (P4, R, i, c).

We obtain 0 usable rules (out of 7 rules in the input problem).

Processor output: { D6 = (P4, {}, i, c) }.

***** We apply the Reduction Pair [with HORPO] Processor on D5 = (P3, R2, i, c).

Constrained HORPO yields:

  skipfold#(F, n, k, cons(x, ys)) (>) skipfold#(F, n, k, skip(k - 1, ys)) | k = k
  skip(k, nil) (>=) nil | k = k
  skip(k, ys) (>=) ys | k <= 0
  skip(k, cons(x, ys)) (>=) skip(k - 1, ys) | k > 0

We do this using the following settings:

* Disregarded arguments:

  cons      1 
  skipfold# 2 

* Precedence and permutation:

    cons       { }   2
  = skipfold#  { 1 } _ 4 _ 3
  > nil        { }
  = skip       { 2 } 1

* Well-founded theory orderings:

  [>]_{Bool} = {(true,false)}
  [>]_{Int}  = {(x,y) | x > -1000 /\ x > y }

All dependency pairs were removed.

Processor output: { }.

***** We apply the Reduction Pair [with HORPO] Processor on D6 = (P4, {}, i, c).

Constrained HORPO yields:

  zip#(cons(x, xs), ys) (>) zip#(ys, xs)

We do this using the following settings:

* Disregarded arguments:

  cons 1 

* Precedence and permutation:

    cons  { 2 }
  > zip#  { 1 2 }

* Well-founded theory orderings:

  [>]_{Bool} = {(true,false)}
  [>]_{Int}  = {(x,y) | x < 1000 /\ x < y }

All dependency pairs were removed.

Processor output: { }.